Answer:
The correct answer is 24
Step-by-step explanation:
to solve this you will need to use the pathagreom theorum
a^{2}+b^{2}=c^{2}
A= one side lenth
B= the secons side lenth
C= hypotnuse
It is helpfull to draw out the situation
you know that the latter is 25 ft, that is your hypotnuse
you also know that the 7 ft away from the base of the building is one of the side lenths, lets call it side a
so plug the numbers into the equation
7^2 + b^2 = 25 ^2
you leave b^2 alone because that is the side you are trying to find
now square 7 and 25 but leave b^2 alone
49 + b^2 = 625
now subtract 49 from both sides
b^2 = 576
now to get rid of the square of b you have to do the opposite and square root both sides removing the square of the B and giving you the answer of..........
B= 24
Hope this helped!! I tryed to explain it as simpil as possiable
<span>280
I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself.
You can choose 1 of 7 appetizers. So we have
n = 7
After that, you chose an entre, so the number of possible meals to this point is
n = 7 * 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 * 4 = 280
Therefore the number of possible meals you can have is 280.
Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 * 1010 * 44 = 3421880
But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>
just subsitute
f(4)=4²+3(4)-1
f(4)=16+12-1
f(4)=28-1
f(4)=27
it approaches 27 as x approaches 4
The answer is 4526
Hope this helps!
